$ \int_1^4 \int_1^{\sqrt{x}} dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_1^2 \int_{y^2}^4 dx \, dy$ (Choice B) B $ \int_1^4 \int_2^{y^2} dx \, dy$ (Choice C) C $ \int_0^2 \int_1^{y^2} dx \, dy$ (Choice D) D $ \int_1^4 \int_{y^2}^2 dx \, dy$
Solution: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $1 < x < 4$ and $1 < y < \sqrt{x}$. Therefore: ${2}$ ${4}$ ${2}$ ${4}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $1 < y < 2$. Then we can define $x$ in terms of $y$. Thus, $y^2 < x < 4$. In conclusion, the double integral after switching bounds is: $ \int_1^2 \int_{y^2}^4 dx \, dy$